Statement: Suppose $f:X\rightarrow Y$ is a diffeomorphism of manifolds with boundary. Show that $\partial f:\partial X\rightarrow \partial Y$ is a diffeomorphism.

Proof: First, let's verify that $f(\partial X)\subset \partial Y$ . Let $x \in \partial X$; $y = f(x) \notin \partial Y$ . Let $\psi : V\rightarrow U$ be a local parameterization of an open neighborhood of $y$ in $Y$ . Then
$f^{-1}\circ \psi : V\rightarrow f^{-1}(U)$ is a diffeomorphism to an open neighborhood of $x$ in $X$.
Hence $x$ is not in $\partial X$. Thus $f(\partial X) \subset \partial Y$ . By applying same arguments to $f^{-1}$ we conclude that $f(\partial X) = \partial Y$. Thus $\partial f : \partial X\rightarrow \partial Y$ is
bijective (it is onto and injective since f is). $\partial f$ is smooth as the restriction
of a smooth $f$ and its inverse is also smooth (again as the restriction of a smooth $f^{-1}$).

So I don't understand how $f^{-1}\circ \psi : V\rightarrow f^{-1}(U)$ is even defined because $\psi: V\rightarrow U$ and $f^{-1}:Y\rightarrow X$. Also, the conclusion that $x$ is not in $\partial X$ is arrived at because $x$ cannot be on the boundary and have an open neighborhood around it diffeomorphically mapped to an open nbd of $\partial Y$, am I correct? Thanks and appreciate a hint.

1 Answer
1

We have $U\subset Y$. Whenever $g:A\to B$ and $B\subset C$ and $h:C\to D$, when we write $h\circ g$, we really mean $h|_{B}\circ g$. And if $p:A\to C$ and $p(A) \subset B$, then we also often write $p$ for the function $p:A\to B$ which matches the other $p$ in the obvious way.

We are assuming $y\notin \partial Y$, so that $y$ has a neighborhood $U$ diffeomorphic to an open subset of $\mathbb{R}^n$ for some $n$. And $x$ has no such neighborhood, being a boundary point. But $f^{-1}(U)$ is such a neighborhood.

$\begingroup$Thanks, I understand why $x$ is not in $\partial X$ now, but for the first paragraph, $\psi:V\rightarrow U$, $f^{-1}:Y\rightarrow X$ and $U\not\subset Y$, right? in fact $U\subset X$ which is confusing me?$\endgroup$
– manifoldedFeb 28 at 16:35

$\begingroup$No, $U \subset Y$ is a neighborhood of $y$ in $Y$ and $V \subset \mathbb{R}^n$ is open with $\psi:V\to U$ a chart for $U$.$\endgroup$
– csprunFeb 28 at 16:37